Hacker News Reader

Comment by johncarlosbaez | Hacker News Reader

It's too bad the top-rated answer on Reddit says "GR does not say Gravity is not a force (or if you do say it’s not a force, then none of the other forces are forces either)" rather than explaining what people mean when they say gravity is not a force (basically, it affects the geometry of spacetime in such a way that an unaccelerated particle can still move along a path that's not a straight line in the traditional sense) and why nonetheless we can treat gravity approximately (that is, perturbatively) as if it were a force, and why this perturbative description when quantized predicts a spin-2 boson, the graviton. Oh well.

View on HN

I had a prof explain it using basic analogy. Forces are things that can move other things. An electrical field can move a metal ball. So too can a metal bat. put a g-meter on the ball and it will register acceleration as force is applied to the ball. Gravity is different. Put a g-meter on a falling metal ball and it will detect zero acceleration. No force is acting upon a ball. Put that metal ball in an eccentric orbit around the earth and it will speed up and slow down during each orbit, but the g-meter will register zero acceleration. The ball falls but is no not accelerated. So gravity is not a force because it doesn't move things. Gravity is something different.

But the acceleration meter won't measure any force because gravity is acting on every part of it uniformly. If you had an acceleration meter entirely made out of the same magnetic substance, and you brought a magnet near it, would the acceleration meter register anything, or would it read zero acceleration, since all parts of it were being acted on uniformly (and thus didn't "notice" any acceleration)?

(Replying to myself)

I guess one answer to this is that particles which are (supposedly?) massless, like the photon, are affected by the space-time warping effects of gravity. A parallel construction wouldn't be true of magnetism or an electric field. Furthermore, when we detect gravity waves, they come at the same time as corresponding gamma ray bursts; since the gamma rays are affected by the space-time warping effects of gravity, this means that the gravity waves themselves are affected by the space-time warping effects of gravity.

So gravity probably is something else. But who knows!

That could also be explained by light having momentum which is affected by (but does not generate) the force of gravity acts like mass.

> the acceleration meter won't measure any force because gravity is acting on every part of it uniformly

No. The acceleration meter won't measure anything because there is nothing to measure. An object in free fall is in free fall; there is no "gravity" acting on it at all. It's just as if the object were floating out in deep space, far from all gravitating bodies. That's the point of the equivalence principle.

> If you had an acceleration meter entirely made out of the same magnetic substance, and you brought a magnet near it, would the acceleration meter register anything

Yes. Electromagnetism, weak, and strong interactions all make the acceleration meter register nonzero, even if the act "equally" on all parts of an object.

In other words, there is no gravity field, in the way there is an EM field that propagates these forces. Or, the "gravity field" is the fabric of space-time itself

Yes.

> The acceleration meter won't measure anything because there is nothing to measure.

Put this way, isn't it almost begging the question? In GR the definition of acceleration is movement in contrast with the movement of gravity. If course gravity will never meet this criteria - all movement due to gravity will be aligned with movement due to gravity.

If instead we had a universe where instead of all matter having a gravitational effect, it was that all matter had a magnetic effect the we'd see no acceleration due to the magnetic effect and gravity would "produce a field" and cause acceleration in your above examples.

You can't use a gravitational biased tool to proclaim gravity is a neutral actor and everything else is a field.

It seems like more accurately, everything is "gravitationally charged", so instead we say it warps spacetime, but really is no different.

> Put this way, isn't it almost begging the question? In GR the definition of acceleration is movement in contrast with the movement of gravity.

No, it isn't. You have it backwards. The definition of acceleration in GR is proper acceleration, i.e., what an accelerometer reads. The "movement" property is then a consequence of this plus picking an appropriate frame of reference.

> If instead we had a universe where instead of all matter having a gravitational effect, it was that all matter had a magnetic effect the we'd see no acceleration due to the magnetic effect

Yes, you would, because unlike gravity, magnetism does not obey the equivalence principle, so differently charged objects in the same magnetic field with the same initial conditions can have different motions. With gravity, all objects in the same field with the same initial conditions have the same motion, regardless of their mass. That is why it is possible to model gravity using spacetime curvature, and that property is unique to gravity.

I'm sure you are familiar with Kalusa-Klein theory (knowing that it is incomplete and/or doesn't describe our universe), don't you think that motion in that theory is less bound to gravity alone like the parent comment suggests because the notion of proper time is different?

I tend to gravitate toward the same line of thinking because of the existence of black hole charge limit.

You might be interested in https://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory

> Yes. Electromagnetism, weak, and strong interactions all make the acceleration meter register nonzero, even if the act "equally" on all parts of an object.

I’m genuinely curious how that acceleration meter would work. There won’t be any internal forces as a consequence of the external field and no relative motion.

> I’m genuinely curious how that acceleration meter would work.

Look up how the one in your phone works. It reads nonzero when you are standing on Earth because of electromagnetic repulsion between your atoms and the atoms in the floor.

> There won’t be any internal forces

Yes, there will, because the object's internal state (including its shape, size, and internal stresses) when it is accelerated is different from its shape when it is in free fall. Why? Because the acceleration sets up internal forces in the object that result in a different equilibrium from the one it was in while it was freely falling.

> It reads nonzero when you are standing on Earth because of electromagnetic repulsion between your atoms and the atoms in the floor.

It works because there's an external force pushing on the surface of the phone, and not equally on all its parts, which is the scenario we are discussing.

> Because the acceleration sets up internal forces in the object that result in a different equilibrium from the one it was in while it was freely falling.

And the cause of this is what you fail to explain.

I think the burden of explanation is on the great great grandparent post which proposed an accelerometer "entirely made out of the same magnetic field" which would let it satisfy an equivalence principle for "magnetic force".

> It works because there's an external force pushing on the surface of the phone

Which then gets transmitted by the surface of the phone to the rest of the phone.

> not equally on all its parts

Yes, equally on all its parts, once you take into account that the parts of the phone can exert forces on each other.

> the cause of this is what you fail to explain

Cause of what? The internal forces? That's obvious: the distances between adjacent atoms change, and the electromagnetic forces between adjacent atoms are distance dependent. In other words, as I have already pointed out, the size and shape of the object changes when it is accelerated (in the sense of proper acceleration), compared to when it is not. Or, to put it another way, the equilibrium state of the object is different when it is accelerated than when it is not, with different inter-atomic distances and therefore different internal forces (as in, nonzero internal forces when accelerated, compared to zero when not).

> Which then gets transmitted by the surface of the phone to the rest of the phone.

But this is a very different situation than when you have an external force acting equally on all the atoms equally, and you will, as you correctly point out, have forces between the atoms. Of course in this case it is trivial, because you can easily measure the tension between different parts (for example).

But that is not the situation we are discussing. The situation we are discussing is when were are an external force on all parts equally. So there will be no tension that you can measure.

> this is a very different situation than when you have an external force acting equally on all the atoms

Gravity is not an "external force" in GR. In GR, an object moving solely under gravity, i.e., in free fall, feels no force (no internal stresses, zero reading on an accelerometer) because there is no force--not because there is "an external force acting equally on all the atoms".

If you want to say that the Newtonian interpretation, where gravity is "an external force acting equally on all the atoms", is indistinguishable experimentally from the GR interpretation, where gravity isn't a force at all, I suppose that's true. But then the "external force acting equally on all the atoms" is just like Carl Sagan's undetectable dragon in his garage. We have a model that works just as well without it, so it gets scraped right off by Occam's Razor. That is the GR position.

Also, no matter how you want to resolve the above issue, it still remains true that non-gravitational forces do not obey the equivalence principle, so there is no "spacetime geometry" interpretation for them that works. And those are also the cases where you do have internal stresses in the objects and an accelerometer reads nonzero. So again GR's interpretation--that these cases are forces while gravity is not, and that explains the difference in accelerometer readings--is simpler than the Newtonian one, where you have to argue that gravity is a "force" but doesn't work like other "forces" work.

Put more briefly, gravity does not produce a compressive stress in the phone in free fall. The external force from the table applied to the phone sitting atop it does create compressive stress. Ergo, the situations are fundamentally different.

> Electromagnetism, weak, and strong interactions all make the acceleration meter register nonzero, even if the act "equally" on all parts of an object.

I'm struggling to wrap my head around this assertion. If all parts of the object are acted upon "equally" (why is this in quotes?) where would this acceleration come from?

> If all parts of the object are acted upon "equally" (why is this in quotes?) where would this acceleration come from?

It is proper acceleration, not coordinate acceleration. An object can have nonzero proper acceleration even if none of its parts are in relative motion. Geometrically, proper acceleration corresponds to path curvature of the worldlines of the atoms in the object. "No relative motion" means all the worldlines of the object's atoms have the same path curvature (modulo some technicalities that don't really matter here). It does not require that that path curvature be zero.

Physically, a typical accelerometer works by measuring the internal stresses that are set up in an object when it is accelerated. These stresses put the object into a different equilibrium state than it was in when it was freely falling: the object's size and shape can change. For typical solid objects at typical Earthbound accelerations these changes are too small for us to see with our unaided senses--but sensitive instruments like accelerometers can detect them.

But that path curvature of worldlines is in spacetime, the field of gravity. If you chose your worldline along the electromagnetic field instead, you'd see no "proper" acceleration when being pulled by the electromagnetic field and you would see "proper" acceleration due to gravity.

And to develop internal stresses, there needs to be some difference in the forces acting upon different parts of the body. Again, you are arbitrarily using free fall as your choice of a default state, against which you are comparing. If you instead chose being stationary on the ground, which would correspond to when you are in electromagnetic free fall, you would register an acceleration when in gravitational free fall.

> that path curvature of worldlines is in spacetime

It's the path curvature of worldlines in spacetime.

> the field of gravity

Only if the spacetime is curved. But worldlines can have path curvature even in flat spacetime.

> If you chose your worldline along the electromagnetic field instead, you'd see no "proper" acceleration when being pulled by the electromagnetic field and you would see "proper" acceleration due to gravity.

There is no such thing as "worldline along the electromagnetic field" as you appear to be using the term. A charged object in an electromagnetic field will have nonzero proper acceleration, as measured by an accelerometer. That's an invariant prediction; there is no alternate model in which it's any different.

> to develop internal stresses, there needs to be some difference in the forces acting upon different parts of the body

No, there doesn't. There just needs to be a difference in the shape and size of the body, from its state when under no external forces.

It is true that, if the body is large enough, the distribution of internal stresses in the body might not be uniform when the body reaches its equilibrium state under some externally applied force. For example, if the object is tall enough, the internal pressure at the top will be measurably less than the internal pressure at the bottom. But this is not due to any difference in external force being applied to the object. It's due to how the object adjusts itself to be in equilibrium under the applied external force.

> If you instead chose being stationary on the ground, which would correspond to when you are in electromagnetic free fall, you would register an acceleration when in gravitational free fall.

Again, this is just wrong. There is neither a valid theoretical model, nor any experimental data, to support this claim.

You are grossly misunderstanding what I am saying, and I don't know how better to explain it.

As far as I can tell, you are claiming that there is a viable model in which, for example, a charged object moving in an electromagnetic field would have "zero proper acceleration" while an uncharged object moving solely under gravity, would have "nonzero proper acceleration". But such a model can't be right, because "proper acceleration" is a direct observable, and our direct observation is the other way around.

It is so difficult to remotely understand all this when every other term is in "quotes". Of course, I'm not a physicist, so I probably wouldn't understand it all, anyways. I am just pointing out how many dang quotes there are. Could you guys maybe define the terms first so you don't have to put two fingers around every concept?

BTW I made this "comment" by "clicking" on the reply button and then I "typed" it using my "keyboard". No, I don't mean "keyboard" when I say "keyboard", hence the quotes.

Inertia causes the sensor in the accelerometer to read nonzero.

Meta: let's stop arguing what might happen, and do the experiment!

> Inertia causes the sensor in the accelerometer to read nonzero.

This can't be right, because an object moving solely under gravity is moving solely under its own inertia, but an accelerometer attached to it reads zero.

> do the experiment!

Do what experiment? Experiments showing that accelerometers attached to objects moving solely under gravity read zero, while objects subjected to non-gravitational forces read nonzero, have been done.

Yeah, if every part of human body is accelerated equally, there's no way to feel anything (you could see it). No matter how hard you get hit or thrown around, you can't measure it internally. Each cell can only feel their neighbors, and there's no internal stress / force between cells.

This isn’t true. Try jumping off a building and you’ll find the acceleration to be quite noticeable ;)

Only because you feel the air rushing past, and the impact with the ground.

If you do this in a capsule where the air is moving with you, and avoid hitting the ground, we call the sensation "weightlessness" or "zero g", like is experienced by astronauts in orbit.

Gravity is absolutely acting on astronauts orbiting the earth, at nearly the same strength as if they were standing on the ground. Depending on the shape of the orbit their linear speed may be increasing or decreasing, and they are definitely experiencing directional acceleration as their path bends in a circle around the Earth. But internally there is no bodily sensation of acceleration. It feels the same as floating, or free fall without the air rushing past.

Yes but the feeling of weightlessness is something you can distinguish!

But you don't feel the force of acceleration. Floating weightless and motionless in space feels the same as being in an elliptical orbit around a planet.

Even more, accelerating due to a being in an (idealized) rocket also feels the same.

https://www.npl.washington.edu/eotwash/equivalence-principle

This has to be true, because if you can't (internally) detect the difference between nothing and a planet being nearby, you obviously also can't detect how massive the planet is, so you can't know if your acceleration relative to the nearby environment is due to gravity or something else, or even how much "absolutel" acceleration you have.

Standing on earth (or the floor of your rocket) feels different due to electromagnetic effects of the nearby "touching" external objects.

I did sky diving some years back. I did feel acceleration immediately after the jump for a second, but then afterwards it was as if I was lying on bed. So what was that feeling I experienced (it was similar to maintain giant wheel coming down)

Going from standing on a surface where you are resisting the force of gravity (like standing on the ground, or in a stable aircraft) to being in free fall definitely has a sensation.

It's a sensation that actually causes a decent percentage of people to feel physically ill for a period after launching to orbit on a rocket.

It's like going over the top of a hill on a roller coaster where you go from being pulled down into your seat to floating up against the restraints.

But once you're falling, or or in orbit (which is also free fall just outside an atmosphere), you don't feel the changes in velocity (acceleration) due to gravity.

The point is you won't feel it (aside from air resistance) on the way down.

The air too, you only feel because the ground and other air is pushing it upwards

If the impact somehow could stop your all cells at the same rate, you wouldn't feel it. Your body gets compressed because of different rates of acceleration.

If aliens have tech which can apply this type of force field / acceleration, then they won't get squished in their spaceships no matter how hard they accelerate. You basically need a large force field like gravity, instead of transferring force via small intermolecular force fields.

This is a clever analogy, but it's actually a little specious.

The reason the "g-meter" (e.g. a weight on a scale) doesn't move in the gravity case is that the weight is affected by the same field. The weight is a weight and feels gravity just like you and everything else in your environment does.

But by construction, you're imagining that the scale you have holds a different electrical charge than the object to which it's attached. Which is "normal" according to our everyday experience, but just an artifact of the way charges work on large objects (they distribute themselves on the "outside" of a conductive environment and everything inside tends to have a neutral distribution).

But that's just arbitrary. You could equally demand (in your gedankenexperiment, though doing this in practice would be very difficult) that your electrical charge be distributed just like the mass is, in which case the force measured would be zero too.

As an aside there are real gravimeters that aren’t scales and they are sensitive enough to detect a person walking around the room they are in and snow accumulating on that building’s roof. Yes that’s right they detect the gravitational force exerted by the snow’s mass.

> Yes that’s right they detect the gravitational force exerted by the snow’s mass.

No, they don't. Gravimeters of the type you describe measure the coordinate acceleration of a freely falling test object in the accelerated frame of the gravimeter. In other words, it's the gravimeter (the part that isn't the freely falling test object) that has a force acting on it, which makes it accelerate upward (proper acceleration--an accelerometer attached to the gravimeter reads nonzero), and a freely falling test mass therefore appears to accelerate downward (coordinate acceleration in the frame of the gravimeter), just as if the gravimeter were inside an accelerating rocket out in deep space far from all gravitating bodies.

In other words, gravimeters of this type rely on the equivalence principle, which is the same principle that GR uses to justify the statement that gravity is not a force.

Rather more pertinently to the question "why is gravity different?", we can measure the time dilation caused by a gravitational potential change of less than 1cm near the surface of the Earth: https://physicsworld.com/a/gravitational-time-dilation-measu...

> put a g-meter on the ball and it will register acceleration as force is applied to the ball. Gravity is different. Put a g-meter on a falling metal ball and it will detect zero acceleration

Put an electrical field generator on a falling metal ball and it will detect changing electric field though...

Well, a changing electric field is not a force tho.

Gravity in GR is a 'fictitious force' or an apparent force, meaning it is is an artifact of your chosen frame of reference.

A feather falling the same speed as an elephant in the Earth reference frame is an example.

It is the same as other apperant forces like centrifugal force.

Even Newton himself said 'Hypotheses non fingo,' or 'I feign no hypotheses. '

It was a conjecture.

> So too can a metal bat. put a g-meter on the ball and it will register acceleration as force is applied to the ball. Gravity is different. Put a g-meter on a falling metal ball and it will detect zero acceleration.

That really depends on construction of your g-meter. If instead of mass (ie. gravitational charge) you use electric charge in your accelerometer then that electric charge on the falling ball - ie. moving with acceleration - will generate EM wave thus providing clear detection of acceleration.

Wrt. the "boson" - gravity effects propagate with finite speed, i.e. wave, and the neutron in gravitational potential experiment shows that the gravitational potential/energy is quantized, and thus we have wave and quantized nature -> boson (wave packet/quant mediating interaction of a charge with the field).

Your charge-based g-meter would also measure zero acceleration in free fall.

https://en.wikipedia.org/wiki/Paradox_of_radiation_of_charge...

> the radiation goes into a region of spacetime inaccessible to the co-accelerating, supported observer. In effect, a uniformly accelerated observer has an event horizon, and there are regions of spacetime inaccessible to this observer

Curious interpretation, but beware this bit wasn't substantiated that well.

Yeah this never made sense to me. Yes, it detects no acceleration. Great. But we know it IS accelerating; if you fall your speed will increase, this can only occur if you are accelerating?!

How would you measure this alleged speed increase, internally?

You can't.

I sure can’t. But why would I want to measure speed increase internally, when it’s due to an external field?

Quick question, how would I even measure speed itself internally? I thought motion and speed was always measured relative to something else, why would an increase of these properties then have to be internal?

Not trying to be a dick, I’m genuinely curious. And yes, I obviously do not know anything about physics. Any explanation or link to a source to help me understand?

I think that more generally the answer is not responsive to the actual question, which is: why do GR and QM need to be unified? To be fair, the question itself wrongly conflates "gravity needs a gauge boson" with the question about GR and QM being unified. Not all theories of quantum gravity involve a "gauge boson" for gravity along the lines of the other Standard Model interactions.

Nor does the basic rationale for why we think gravity needs to be quantized involve a "gauge boson". It involves simple reasoning about how QM works. Say we have an experiment which puts an object with non-negligible stress-energy into a superposition of being in two different positions (for example, we make its position depend on the outcome of a spin measurement on a qubit). QM would say that spacetime would then need to also be in a superposition of two different geometries. But GR, as a classical theory, has no way to handle that. We would need a quantum theory of gravity, i.e., a quantum theory that can handle superpositions of different spacetime geometries.

>Say we have an experiment which puts an object with non-negligible stress-energy into a superposition of being in two different positions (for example, we make its position depend on the outcome of a spin measurement on a qubit). QM would say that spacetime would then need to also be in a superposition of two different geometries.

The energy to move the object into a given position is an additional element here unaccounted for in your model. 2 different positions to move object into - 2 different energies (more specifically 2 different changes to the starting, before the experiment, stress energy distribution of the Universe). When corresponding moving energies (ie. their GR effects) are accounted for in those 2 cases it may as well be that those 2 cases are indistinguishable from the GR point of view, ie. those 2 supposedly different spacetime geometries happen to be the same. The superposition of 2 indistinguishable cases - it doesn't really matter is it superposition or not.

There is an entire sub-field of experimental science that involves putting ever larger objects in superpositions of different locations. These experiments are no closer to testing quantum gravity, but they falsify whatever it is you're trying to say here. See https://www.nature.com/articles/srep13884 for a random example.

> The energy to move the object into a given position is an additional element here unaccounted for in your model.

You can set the experiment up so the energy is the same in both cases (for example, both positions at the same height, just horizontally separated). If you don't, then yes, you have to include the effects of the different energies in your model.

> When corresponding moving energies (ie. their GR effects) are accounted for in those 2 cases it may as well be that those 2 cases are indistinguishable from the GR point of view, ie. those 2 supposedly different spacetime geometries happen to be the same.

I'm not sure how this would work if the energies were different, since "different" means a different source for the spacetime geometry.

But in any case, yes, for such an experiment to be relevant at all to the question I was discussing, the spacetime geometries being superposed would have to be different.

>You can set the experiment up so the energy is the same in both cases (for example, both positions at the same height, just horizontally separated)

The action of placing them, say with your hands for simplicity, into different horizontal positions means differently pushing the Earth with your legs. For more cleaner illustration - let's say in our experiment a space ship is placed into orbit clockwise or anti clockwise. We can't just teleport the ship, so let's say we move it by rocket engines. So the ship goes in one direction, rocket engine exhaust goes in the other. The exhaust does have mass and speed. Even if it wouldn't eliminate the superposition gap, it will definitely decrease it, and decreasing the superposition gap increases the chances that some other unaccounted for factor(s) (for example gravitational waves caused by all these movements) will eliminate it or decrease further. Even if ultimately we still can't fully eliminate the gap, significantly decreasing it may eliminate various divergencies arising from quantization or make them very smallscale/localized (an observer from Alfa Centauri wouldn't care about the ship's orbit direction like we don't care about the spin of a given particle in the air around us) and average-able out on larger scales.

> The action of placing them, say with your hands for simplicity, into different horizontal positions means differently pushing the Earth with your legs.

This might change the momentum, but not the energy if the heights are the same. But if your point is that there will always be some difference in a conserved quantity, yes, that's a fair point.

But it also means that there will always be some difference in the spacetime geometry. None of the other factors you talk about would eliminate the "superposition gap", because none of them cancel out any changes in the spacetime geometry; they just add more changes to it.

> average-able out on larger scales

But if you don't have a theory that can represent the variations you're going to average out, you can't do the averaging. That's the problem: classical GR cannot represent "variation in spacetime geometry" at all. It can only represent one spacetime geometry. It can't represent a superposition of them, not even to do an average.

I would imagine the desire to unify GR and QM is because if we expect the universe to operate on a set of rules, they should by unified. And we just need to find out how.

But the forces and factors that work on the smallest particle should be the same forces that work on the largest of galaxies. If they are not, that's a completely different mystery and means our entire view of the universe is missing something substantial.

> I would imagine the desire to unify GR and QM is because if we expect the universe to operate on a set of rules, they should by unified.

That is one key principle that drives the effort, yes. However, that doesn't mean things will always work out that way. Freeman Dyson, for one, published at least one paper making arguments for why gravity didn't need to be quantized.

> the forces and factors that work on the smallest particle should be the same forces that work on the largest of galaxies.

If you mean fundamental forces, then this is true (that's the definition of "fundamental"), but it also means that you have to adopt many levels of indirection between those fundamental forces and what actually happens with macroscopic objects. Or, to put it another way, the models we actually use to make predictions can have "forces and factors" in them that are not any of the fundamental ones, and that's fine, as long as we have some chain of reasoning that connects those models to the fundamental forces and factors. For example, our models of macroscopic objects can have dissipative forces like friction and viscosity in them; those aren't fundamental forces. But we have a chain of reasoning that connects them to fundamental forces (electromagnetic forces between electrons in atoms).

My layperson understanding is that GR specifies that energy/mass-momemtum influences space-time curvature which then produces gravity (as particles travelling along straight lines in the curved space). That would imply that fermions are therefore the force carriers for gravity, not a hypothetical new boson.

Well, bosons also have mass and can distort space. Theoretically even the (rest-)massless photon distorts spacetime, think it's called a kugelblitz. Also how would fermions be the force carriers if they don't physically move through space themselves to interact with far away fermions, ex gravitational waves? Unless you're advocating for a relational model of space which hey I'm all for but introduces other issues afaik

> My layperson understanding is that GR specifies that energy/mass-momemtum influences space-time curvature which then produces gravity (as particles travelling along straight lines in the curved space).

That's correct. However...

> That would imply that fermions are therefore the force carriers for gravity, not a hypothetical new boson.

That's wrong. Energy/mass-momentum (which can, as hughesjj points out, be bosons or fermions or both) is the source of gravity. The source is not the same as the "force carrier". (For example, in electromagnetism the source is charge/current, but the force carrier is the photon.)

In GR, gravity has no "force carrier" because it is not a force. In the simplest quantum model that has a "force carrier" for gravity, the quantum field theory of a massless spin-2 field, the "force carrier" is the massless spin-2 graviton, which is not the same as any source that occurs in ordinary matter.

What does it mean to 'treat gravity approximately (that is, perturbatively)'? That sounds like something we do to model, which only approximates reality. That model sounds like it shouldn't be used to predict anything else? Or at least whatever is predicted shouldn't be expected to exist in 'reality'.

It’s models all the way down.

John von Neumann: "... the sciences do not try to explain, they hardly even try to interpret, they mainly make models. By a model is meant a mathematical construct which, with the addition of certain verbal interpretations, describes observed phenomena. The justification of such a mathematical construct is solely and precisely that it is expected to work—that is, correctly to describe phenomena from a reasonably wide area."

Both GR and QFT are insanely accurate models, but they are just models.

The N-body problem is undecidable, and Gödel, Turing, Church and other s proved that is the best we can do.

https://philsci-archive.pitt.edu/13175/1/parker2003.pdf

Western reductionism or Laplacian determinism is a good framework for practical, computable models. QFT actually is actually one of the counterexamples to Western reductionism.

But models are reductive and scientific models are just models. Don't confuse the map for the territory.

All models are wrong, some are useful; is another way of saying the same thing.

Could you elaborate or share some articles on why QFT is a counterexample to western reductionism ? In laymans terms that is.

The oversimplified version is Laplac's deamon can't split quantum superposition.

Superposition being inseparable is a large part on why the many words concept is popular with some people. It is about regaining a form of determinism.

There isn't a model out there that doesn't break down at some point.

Looking at gravity, we can compare what General Relativity and Quantum Mechanics say about the center of a black hole. Remember, both GR and QM have both been able to accurately model the way the world works at every scale we have been able to measure and test them. But they are incompatible with each other in certain points, such as the singularity in the black hole. GR says the center of the black hole is an infinitely dense point. QM says this can't be true because everything is made up of waves in a field, which requires things be spread out over some amount of an area. These can't both be true, yet GR and QM have both stood up to every single test and observation we can throw at them. Every prediction they make that we can verify has been verified and lines up with the theories. And this is not the only place they disagree, of course, but it is one example.

And that's really, from my understanding, the more fundamental answer to the question asked in the reddit post. It's not that unifying the two requires a gauge boson like the graviton, though that it is one possible outcome when quantizing gravity, but that we have two very useful and very tested models of how things work that are incompatible with each other in certain ways. Maybe gravitons exist, though it currently seems impossible for us to reach the point where we can detect them - Dyson calculated that using an Earth sized detector we'd be able to detect about one graviton from the sun per billion years, if they exist - and maybe they don't.

As for predicted things not expecting to exist in reality, this is really just par for the course for models. It's not like tensors are some real physical thing either, for example.

Unaccelerated particle= inertial frame.

Approximately: consider gravity as small deviations from inertiality, so it's approximately linear, like the other forces we know and love.

Q. What's the difference between a boson and a bison?

A. When you incessantly hunt for every boson you can find anywhere, you end up with more of them.

I've recently learned about Kaluza–Klein theories (just curious wiki browsing), do they unify electromagnetism and gravity in the sense that electromagnetism also "works by changing the geometry of spacetime" ? How does this relate to QFT?

> do they unify electromagnetism and gravity in the sense that electromagnetism also "works by changing the geometry of spacetime" ?

Yes, somehow actually. KK theories introduce an extra spatial dimension beyond our usual (3+1) which is postulated to be "compactified". This just means that it's curled up on itself at such a tiny scale that we don't directly perceive it. The way this extra dimension is curled and shaped affects the geometry of the overall 5-D spacetime. How it is connected to EM is that now these geometrical variations in the 5-D spacetime, when viewed from our 4-D perspective, manifest as the EM force and its associated field.

So you can say like in GR which have gravity arises as consequence of geometry, it is in KK that EM is consequence of geometry. However, the geometry and details are different.

> How does this relate to QFT?

Not much in the sense that they can provide useful information to each other. QFT describes forces in terms of interactions mediated by particles (e.g., photons for EM). KK, while primarily geometrical, give hints that perhaps these force-carrying particles can be associated with specific vibrational modes of the extra dimension. Of course KK theory only include EM and gravity. So we know for sure that we need to go beyond KK. This was the actual motivation for people to think about string theory to expand the original KK work.

> KK theories introduce an extra spatial dimension beyond our usual (3+1) which is postulated to be "compactified". This just means that it's curled up on itself

You've just explained "compactified" in terms of being "curled up", but that doesn't really help (for me, at least). What does it mean for a dimension to be "curled up"?

I'm sorry if I'm confusing. That's the term I use, usually assuming that it is obvious (like referring to C++ templates to a python developer without explaining what is that).

To be honest, it is hard to visualize, as it is counterintuitive of what we think of space. While I know many people would disagree, I really like the garden hose analogy [1]. The idea is to simply imagine a very long garden hose. From a great distance, it looks like a one-dimensional line. Now you get closer, and realize it has a second dimension, which is its circumference curled around that seemingly 1-D line. An ant walking on the hose can move along its length, but also in a circle around it. The extra dimension in KK is like this circumference. It is tiny, curled up so we don't directly notice it, but still potentially there.

[1] https://www.preposterousuniverse.com/blog/2004/06/30/extra-d...

This kind of makes me think of a spatial version of the Time Cube.

Interesting. I ask because I have a suspicion that Quantum theories seem more fundamental than General relativity, because treating geometry within a quantum framework seems very hard and "non-natural", or implausible (e.g. when considering superposition of spacetimes!). While if somehow gravity is a quantum effect (within a simpler space-time framework), that seems much more plausible... but Kaluza-Klein captured my interest in the other direction. Although I'm still thinking the quantum framework is appears to be the correct one, even though the assumptions of GR are very strong (so something like the equivalence principle or some other notable principle needs to break).

The problem with extra dimensions is that such an increase of dimensions preclude pairwise interaction of particles due to lacking a sound relationship between 2-operand dot product, cross product and 2-operand tensor product in those dimensions.

You can say similar things about the field of the other forces too, though. The path of a charged particle in the EM field could be described as that particle experiencing a different space-time geometry, arising from the EM and gravitational field combined, and thus EM could also be seen as a geometry and not a force. In fact, the impulse of photons, which are vibrations in the EM field, does affect the curvature of space-time, similar to how particles with mass do.

> The path of a charged particle in the EM field could be described as that particle experiencing a different space-time geometry

No, it can't, because the EM interaction does not obey the equivalence principle, as gravity does. The geometric interpretation of gravity relies on the equivalence principle.

To state this another way: if I put two objects with different masses at a given point in spacetime and give them both the same initial velocity in the same direction, their paths through spacetime under gravity will be the same. But if I put two objects with different charges at a given point in the same electromagnetic field and give them both the same initial velocity in the same direction, their paths through spacetime will not be the same. And this remains true even if I add "extra dimensions" to "spacetime" along the lines of Kaluza-Klein theory, to represent the EM field.

Non gravity forces are quite linear and simple, the coupling with fermions is only through covariant derivative.

Gravity is extremely non linear, when you look at the expression for Ricci tensor, it is much more complicated than other forces.

Help me with the last bit, where the perturbations when quantised predict spin-2 bosons.

Honestly, the flagged/dead answer from ChatGPT makes a lot of sense to me. Can someone explain where it goes wrong, without resorting to the usual snide remarks about parrots and other whistling-past-the-graveyard rhetoric?

Is it just making up BS regarding the attributes imparted by various spin values, or is that a reasonable explanation of why gravitons are presumed to be spin-2 particles?

Indeed. I shared what I thought was a detailed and helpful generative response to further the discussion and hopefully receive feedback from those knowledgeable on the topic. It's disappointing to see it met with downvotes and flagging, rather than constructive conversation.

Same old story throughout history. People feel threatened by technology, and they lash out.

It's disappointing here, for sure.

Woaw, a top comment by John Carlos Baez himself!

I feel like it's easier to explain that gravity is a force between matter and space(time) and not necessarily between matter and matter, like we presume the other forces are.

I recognize your name and believe you know what you are talking about, so please, please tell us the why! You or someone you know probable has a blog post or article you can link, maybe?

Wow I never look at usernames but maybe I should start. I believe I read his `Gauge Fields, Knots, and Gravity` back in college (or tried to, anyways. it was a touch above my level at the time)

it would be so much easier to visualize it if we could imagine 4th dimension.

Reply on news.ycombinator.com